power consumption 45kW

I thought turning up the voltage to achieve the specified kV/mm
was what you meant by 'max value' in the example section.

Harry

Michel Jullian wrote:

> Better: area OK, but power still wrong.
> 
> Michel
> 
> ----- Original Message -----
> From: "Harry Veeder" <[EMAIL PROTECTED]>
> To: <[email protected]>
> Sent: Monday, February 26, 2007 11:07 AM
> Subject: Re: [Vo]: New multiwire-plane lifter design guide (was Re: lifter in
> a accelerating frame)
> 
> 
>> Michel Jullian wrote:
>> 
>>> 
>>> Harry wrote:
>>> 
>>>>> 100 kg thrust, 1 m gap, 0.9 kV/mm
>>> ...
>>>> Is this correct?
>>>> 
>>>> power consumption  9 kW
>>>> area               3.1 m^2
>>>> wire spacing       1.3m
>>> 
>>> Only the wire spacing is correct, check your power and area calculations.
>> 
>> 
>> power consumption  90 kW
>> area               310 m^2
>> 
>> Harry
>> 
>>>> Suggestions:
>>>> Rephrase the scaling rules to read:
>>>> 
>>>> "The W per g are proportional to the kV/mm, whereas the m^2 per g are
>>>> proportional to the inverse square of the kV/mm."
>>>> 
>>>> I would also explicitly show the mass in the calculation
>>>> of the area in the example section.
>>> 
>>> Thanks for the improvements, revised version:
>>> *********************************************
>>> MULTIWIRE-PLANE LIFTER/IONOCRAFT DESIGN GUIDE:
>>> --------
>>> Reference design: At 1 kV/mm (the max we can do without arcing) the power
>>> consumption is 2 W per "gram" of thrust, and the required area is 0.0025 m^2
>>> per g.
>>> 
>>> Scaling rules: The W per g are proportional to the kV/mm, whereas the m^2
>>> per
>>> g are proportional to the inverse square of the kV/mm.
>>> 
>>> Wire: as thin as possible (0.1mm OK in most cases), wire-wire spacing = 1.3
>>> times the gap length d (optimum value)
>>> 
>>> EXAMPLE:
>>> ---------------
>>> We want to lift 50 g, and we choose a v/d of half the max value i.e. 0.5
>>> kV/mm, namely v=25kV for a d=50mm gap, to save on power consumption (our
>>> color
>>> monitor is only 75W)
>>> 
>>> 1/ Required power per g: 2 W * 0.5 = 1 W  -> Power consumption for 50 g = 50
>>> W
>>> 2/ Required area per g: 0.0025 m^2 / 0.5^2 = 0.01 m^2 -> Area for 50 g =
>>> 0.50
>>> m^2
>>> 3/ Wire-wire spacing: 1.3*50 mm = 65mm
>>> 
>>> *********************************************
>>> --
>>> Michel

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